`=>` Let `R, S` and `T` be three non collinear points on the plane with position vectors `vec a ,vec b` and `vec c` respectively (Fig).
`=>` The vectors `bar (RS) ` and `vec (RT)` are in the given plane. Therefore, the vector `bar (RS) xx vec (RT)` is perpendicular to the plane containing points R, S and T. Let `vec r` be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector `bar (RS) xx bar (RT) ` is
`(vec r - vec a) * (bar (RS) xx bar (RT) ) = 0`
or `color{red}{( vec r - vec a) xx [ (vec b - vec a) xx (vec c - vec a) ] = 0}` ........(1)
`=>` This is the equation of the plane in vector form passing through three noncollinear points.
`color{red} " Key Point"`- It's necessary to say that the three points had to be non collinear because If the three points were on the same line, then there will be many planes that will contain them (Fig).
`=>` These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.
`color{brown} "Cartesian form"`
`=>` Let `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)` be the coordinates of the points R, S and T respectively. Let `(x, y, z)` be the coordinates of any point P on the plane with position vector `vec r` . Then
`bar (RP) = (x-x_1)hat i + (y-y_1) hat j + (z-z_1) hat k`
`bar (RS) = (x_2 - x_1) hat i + (y_2 - y_1) hat j + (z_2 - z_1) hat k`
`bar (RT) = (x_3 -x_1) hat i + (y_3- y_1) hat j + (z_3 -z_1) hat k`
`=>` Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have
`\ "Equation of the plane" color{red} { | (x-x_1 , y-y_1, z-z_1 ), ( x_2- x_1 , y_2- y_1, z_2-z_1), ( x_3 -x_1 , y_3 -y_1, z_3-z_1) | =0}`
`=>` which is the equation of the plane in Cartesian form passing through three non collinear points `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)`.
`=>` Let `R, S` and `T` be three non collinear points on the plane with position vectors `vec a ,vec b` and `vec c` respectively (Fig).
`=>` The vectors `bar (RS) ` and `vec (RT)` are in the given plane. Therefore, the vector `bar (RS) xx vec (RT)` is perpendicular to the plane containing points R, S and T. Let `vec r` be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector `bar (RS) xx bar (RT) ` is
`(vec r - vec a) * (bar (RS) xx bar (RT) ) = 0`
or `color{red}{( vec r - vec a) xx [ (vec b - vec a) xx (vec c - vec a) ] = 0}` ........(1)
`=>` This is the equation of the plane in vector form passing through three noncollinear points.
`color{red} " Key Point"`- It's necessary to say that the three points had to be non collinear because If the three points were on the same line, then there will be many planes that will contain them (Fig).
`=>` These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.
`color{brown} "Cartesian form"`
`=>` Let `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)` be the coordinates of the points R, S and T respectively. Let `(x, y, z)` be the coordinates of any point P on the plane with position vector `vec r` . Then
`bar (RP) = (x-x_1)hat i + (y-y_1) hat j + (z-z_1) hat k`
`bar (RS) = (x_2 - x_1) hat i + (y_2 - y_1) hat j + (z_2 - z_1) hat k`
`bar (RT) = (x_3 -x_1) hat i + (y_3- y_1) hat j + (z_3 -z_1) hat k`
`=>` Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have
`\ "Equation of the plane" color{red} { | (x-x_1 , y-y_1, z-z_1 ), ( x_2- x_1 , y_2- y_1, z_2-z_1), ( x_3 -x_1 , y_3 -y_1, z_3-z_1) | =0}`
`=>` which is the equation of the plane in Cartesian form passing through three non collinear points `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)`.